%I #13 May 08 2020 11:51:14
%S 1,1,2,3,6,8,13,18,29,39,57,77,112,148,205,271,372,484,647,838,1110,
%T 1423,1852,2361,3051,3857,4922,6191,7849,9805,12319,15314,19131,23649,
%U 29333,36099,44556,54568,66963,81683,99803,121229,147413,178411,216111,260590,314365,377819,454229
%N Expansion of Product_{k>=1} 1/((1 - x^k)*(1 - x^(4*k))).
%C Convolution of A000041 and A035444.
%C Convolution of A000712 and A082303.
%C Convolution inverse of A107034.
%C Number of partitions of n if there are 2 kinds of parts that are multiples of 4.
%H Zakir Ahmed, Nayandeep Deka Baruah, Manosij Ghosh Dastidar, <a href="https://doi.org/10.1016/j.jnt.2015.05.002">New congruences modulo 5 for the number of 2-color partitions</a>, Journal of Number Theory, Volume 157, December 2015, Pages 184-198.
%H <a href="/index/Par#part">Index entries for sequences related to partitions</a>
%F G.f.: exp(Sum_{k>=1} x^k*(1 + x^k + x^(2*k) + 2*x^(3*k))/(k*(1 - x^(4*k)))).
%F a(n) ~ 5^(3/4) * exp(sqrt(5*n/6)*Pi) / (2^(13/4) * 3^(3/4) * n^(5/4)). - _Vaclav Kotesovec_, Aug 14 2018
%e a(5) = 8 because we have [5], [4, 1], [4', 1], [3, 2], [3, 1, 1], [2, 2, 1], [2, 1, 1, 1] and [1, 1, 1, 1, 1].
%p a:=series(mul(1/((1-x^k)*(1-x^(4*k))),k=1..55),x=0,49): seq(coeff(a,x,n),n=0..48); # _Paolo P. Lava_, Apr 02 2019
%t nmax = 48; CoefficientList[Series[Product[1/((1 - x^k) (1 - x^(4 k))), {k, 1, nmax}], {x, 0, nmax}], x]
%t nmax = 48; CoefficientList[Series[1/(QPochhammer[x] QPochhammer[x^4]), {x, 0, nmax}], x]
%t nmax = 48; CoefficientList[Series[Exp[Sum[x^k (1 + x^k + x^(2 k) + 2*x^(3 k))/(k (1 - x^(4 k))), {k, 1, nmax}]], {x, 0, nmax}], x]
%t Table[Sum[PartitionsP[k] PartitionsP[n - 4 k], {k, 0, n/4}], {n, 0, 48}]
%Y Cf. A000041, A000712, A002512 (self-convolution), A002513, A035444, A082303, A100853, A107034, A318026, A318028.
%K nonn
%O 0,3
%A _Ilya Gutkovskiy_, Aug 13 2018